Brilliant

Let B be the first alphabet and T be the sixth, then there are four words between B and T which can be arranged in $\dfrac{7!}{2! × 2!}$ ,because I and L repeat and there are 7 alphabets other than B and T. The answer is same even if B is second & T is seventh and B is third and T is eighth & B is fourth and T is ninth & T is first and B is sixth &T is second and B is seventh & T is third and B is eighth & T is fourth and B is ninth.

So, the total number of words that can be formed = $\dfrac{7!}{2! × 2!} × 8 = \dfrac{8!}{2! × 2!} = 10080$
$\therefore \dfrac{x}{60} = \dfrac{10080}{60} = \boxed{168}$